Two spherical bodies of masses $M$ and $5 M$ and radii $R$ and $2 R$ respectively are released in free space with initial separation between their centres equal to $12 R$. If they attract each other due to gravitational force only,then the distance covered by the smaller body just before collision, is

• A

  1.2R
• B

  2.5R
• C

  4.5R
• Correct answer
  D

  7.5R

Solution

Let at $O$ there will be a collision. If smaller sphere moves $x$ distance to reach at $O$, then bigger sphere will move a distance of $(9 R-x)$. $\begin{aligned} F & =\frac{G M \times 5 M}{(12 R-x)^2} \\ a_{\text {small }} & =\frac{F}{M}=\frac{G \times 5 M}{(12 R-x)^2} \\ a_{\mathrm{big}} & =\frac{F}{5 M}=\frac{G M}{(12 R-x)^2} \\ x & =\frac{1}{2} a_{\text {small }} t^2=\frac{1}{2} \frac{G \times 5 M}{(12 R-x)^2} t^2 \ldots(i)\\ (9 R-x) & =\frac{1}{2} a_{\text {big }} t^2=\frac{1}{2} \frac{G M}{(12 R-x)^2} t^2\ldots(ii) \end{aligned}$ Thus, dividing Eq. (i) by Eq. (ii), we get $\frac{x}{9 R-x}=5$ $\begin{aligned} & \Rightarrow \quad x=45 R-5 x \\ & \Rightarrow \quad 6 x=45 R \\ & \therefore \quad x=7.5 R \\ & \end{aligned}$

Hence, the correct answer is option: 'D'.